Answer:
x = 5/2 , y = -7/2
Step-by-step explanation:
<em>r's</em><em> your</em><em> solution</em>
<em> </em><em> </em><em>=</em><em>></em><em> </em><em>formula</em><em> </em><em>for</em><em> </em><em>finding</em><em> </em><em>midpoint</em><em> </em><em>=</em><em>.</em><em>(</em><em> </em><em>X1</em><em> </em><em>+</em><em> </em><em>X</em><em>2/</em><em>)</em><em>/</em><em>2</em><em> </em><em>,</em><em> </em><em>(</em><em>Y1+</em><em>Y2)</em><em>/</em><em>2</em>
<em>=</em><em>></em><em> </em><em>putting</em><em> </em><em>the </em><em>value</em><em> </em><em>of </em><em>in </em><em>formula</em>
<em> </em><em> </em><em>=</em><em>></em><em> </em><em>x=</em><em> </em><em> </em><em>4</em><em>+</em><em>1</em><em>/</em><em>2</em><em> </em><em>,</em><em> </em><em>y </em><em>=</em><em> </em><em>-</em><em>1</em><em>-</em><em>6</em><em>/</em><em>2</em>
<em>=</em><em>></em><em> </em><em>x </em><em>=</em><em> </em><em>5</em><em>/</em><em>2</em><em> </em><em>,</em><em> </em><em>y </em><em>=</em><em> </em><em>-</em><em>7</em><em>/</em><em>2</em>
<em>hope</em><em> it</em><em> helps</em>
Answer:
answer what ?
Step-by-step explanation:
Answer:
4:16 or 1:4
Step-by-step explanation:
1/5 of 20 = 4
there are 4 red marbles
there are 16 blue marbles
4:16 or 1:4
5. Given the equation y = (1/4) cos[(2pi/3)*theta]:
5a. For the general equation y = a cos(bx), the period is given by 2pi/b. In this equation, b = 2pi/3, so this means that 2pi/b = 2pi/(2pi/3) = 3. Therefore, the period of this equation is 3, and the cosine wave repeats itself every 3 x-units.
5b. For the general equation y = a cos(bx), the amplitude is given by a. Therefore the amplitude is a = 1/4, and this means that the cosine wave's range is from -1/4 to 1/4 for all values of x.
5c. The equation of the midline is y = 0. This represents the average value over the wave. This is determined by adding the highest and lowest values of the range and taking the average, in this case, 1/4 + (-1/4) = 0, and 0 / 2 = 0. Another way to do this is using the general equation y = a cos(bx) + c, where the midline's equation is y = c. In this case, there is no value of c in the given, implying that c = 0, and the midline is y = 0.
6. Let the horizontal distance be x. Then tan42 = h/x, and h = x tan42. Then using the Pythagorean theorem: 3280^2 = h^2 + x^2
3280^2 = x^2 (tan42)^2 + x^2
3280^2 = x^2 [(tan42)^2 + 1]
x = 2437.52
Therefore, h = x tan42 = 2,194.75 ft.
